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On one exam, the average (arithmetic mean) in a class of 20 students

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On one exam, the average (arithmetic mean) in a class of 20 students  [#permalink]

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New post 04 May 2017, 00:53
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On one exam, the average (arithmetic mean) in a class of 20 students was 80%. A week later, a new student enrolled in the class, and had to make up this exam. If the average on the exam for the class increases to 80.5%, what grade did the new student receive on his exam?

A. 90.5
B. 89
C. 85.5
D. 82
E. 80.5

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Re: On one exam, the average (arithmetic mean) in a class of 20 students  [#permalink]

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New post 04 May 2017, 02:54
Here is a technique I learned from one of the post by VeritasPrepKarishma ma’am :

Mean of 20 People = 80
Mean of 21 People = 80.5
Difference = 0.5 (meaning : addition of one person’s score increases the mean by 0.5)

So the new person will have to contribute = 80.5
Additionally, the new person will have to contribute 0.5 on behalf of the other 20 people to raise the mean by 0.5 ..thus 20 x 0.5 = 10

So total contribution by new people = 10 + 80.5 = 90.5

So, answer = A
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Re: On one exam, the average (arithmetic mean) in a class of 20 students  [#permalink]

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New post 04 May 2017, 05:08
Approach 1:

Let the total score of the test = 100
=> The 20 students scored a total of 20*80 = 1600
When the new student attempted the test, the new average is 80.5.
=> The 21 students scored a total of 21*80.5 = 1690.5
So, the new student scored 1690.5 - 1690 = 90.5

Approach 2:

The new student has to contribute 80.5 (this is the average) + he has to contribute 0.5 more to each of the 20 students = 80.5 + (0.5*20) = 90.5

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Re: On one exam, the average (arithmetic mean) in a class of 20 students  [#permalink]

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New post 04 May 2017, 11:47
My approach

number of students x percentages
20 x 80 = 1600

Then scan through the answer choices. Add 1600 to the answer choice and divide by 21 until you get the answer 80.5

(1600 + 90.5)/2 = 80.5
Answer is A
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Re: On one exam, the average (arithmetic mean) in a class of 20 students  [#permalink]

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New post 06 May 2017, 16:57
Bunuel wrote:
On one exam, the average (arithmetic mean) in a class of 20 students was 80%. A week later, a new student enrolled in the class, and had to make up this exam. If the average on the exam for the class increases to 80.5%, what grade did the new student receive on his exam?

A. 90.5
B. 89
C. 85.5
D. 82
E. 80.5


We are given that the original average for 20 students, on an exam, was 80%. We can use the formula average = sum/number, or equivalently, average x number = sum. Thus, the sum is 80 x 20 = 1600.

We are also given that when a new student enrolls, the average increases to 80.5%. Thus, we can create the following equation in which x = the score of the makeup student:

average = sum/number

80.5 = (1600 + x)/21

1690.5 = 1600 + x

90.5 = x

Answer: A
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Re: On one exam, the average (arithmetic mean) in a class of 20 students  [#permalink]

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New post 07 May 2017, 04:52
Bunuel wrote:
On one exam, the average (arithmetic mean) in a class of 20 students was 80%. A week later, a new student enrolled in the class, and had to make up this exam. If the average on the exam for the class increases to 80.5%, what grade did the new student receive on his exam?

A. 90.5
B. 89
C. 85.5
D. 82
E. 80.5


Earlier total of class = 80*20 = 1600
New total of class = 80.5*21 = 1690.5

The score of new student = The increase in score = 1690.5-1600 = 90.5

Answer: Option A

P.S. Questions of Average can be most easily solves if the total (Average*No. of occurrences) is calculated beforehand
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Re: On one exam, the average (arithmetic mean) in a class of 20 students &nbs [#permalink] 07 May 2017, 04:52
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